Question

A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if

• A. both the length and the radius of the wire are halved
• B. the length of the wire is doubled
• C. the radius of the wire is doubled
• D. both the length and the radius of the wire are doubled

Answer 1

The correct option is D both the length and the radius of the wire are doubled

Heat developed in wire $=\frac{v^2}{R}$

Given that $V=$ constant

$\Rightarrow \propto \frac{1}{R}\propto \frac{\pi r^2}{{\rho }^2}$ since $R=\rho \frac{1}{A}$ and $A=\pi r^2$.

Where $l$ is length and $r$ is radius of wire, $\rho$ is its resistivity

$\Rightarrow$ $H\propto \frac{r^2}{l}$

When $l$ and $r$ both are doubled, then $H^{\prime }\propto \frac{\left(2r^2\right)}{2l}\propto \frac{4r^2}{2l}$

$H^{\prime }\propto 2H$

Hence, option (b) is correct.

Explanation for incorrect answer :

We know that

$\Rightarrow$ $H\propto \frac{r^2}{l}$

Part $\left(a\right)$ When $l$ and $r$ both are halved, then
$H^{\prime }\propto \frac{2(r{)}^2}{4l}\propto \frac{r^2}{2l}$

$H^{\prime }\propto \frac{H}{2}$

Part $\left(c\right)$ When $r$ is doubled, then $H^{\prime }\propto \frac{(2r{)}^2}{l}\propto \frac{4r^2}{l}$

$H^{\prime }\propto 4H$

Part $\left(d\right)$ When $l$ is doubled, then $H^{\prime }\propto \frac{(r{)}^2}{2l}\propto \frac{r^2}{2l}$

$H^{\prime }\propto \frac{H}{2}$

Hence option (a), (c), (d) are incorrect.